Compute the value of \[N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2,\]where the additions and subtractions alternate in pairs.
Answer: Using the difference of squares factorization, we have \[\begin{aligned} N &= (100^2-98^2) + (99^2-97^2) + (96^2-94^2) + (95^2-93^2) + \dots + (4^2-2^2) + (3^2-1^2) \\ &= 2(100 + 98) + 2(99 + 97) + 2(96 + 94) + 2(95 + 93) + \dots + 2(4 + 2) + 2(3+1) \\ &= 2(1 + 2 + \dots + 100) \\ &= 2 \cdot \frac{100 \cdot 101}{2} \\ &= \boxed{10100}.\end{aligned}\]